(2x+10)/((x-3)^2):((x+5)^3)/(x^2)-9

To solve the given expression, we need to simplify it step by step. The expression is:<br /><br />\[ \frac{2x+10}{(x-3)^2} : \frac{(x+5)^3}{x^2-9} \]<br /><br />First, let's rewrite the division as multiplication by the reciprocal:<br /><br />\[ \frac{2x+10}{(x-3)^2} \times \frac{x^2-9}{(x+5)^3} \]<br /><br />Next, factorize the numerators and denominators where possible. Notice that \(2x + 10\) can be factored as \(2(x + 5)\), and \(x^2 - 9\) can be factored as \((x - 3)(x + 3)\):<br /><br />\[ \frac{2(x+5)}{(x-3)^2} \times \frac-3)(x+3)}{(x+5)^3} \]<br /><br />Now, cancel out the common factors in the numerator and the denominator. The term \((x+5)\) appears in both the numerator and the denominator, so we can cancel one instance of it:<br /><br />\[ \frac{2 \cancel{(x+5)}}{(x-3)^2} \times \frac{(x-3)(x+3)}{\cancel{(x+5)}^3} \]<br /><br />This simplifies to:<br /><br />\[ \frac{2}{(x-3)^2} \times \frac{(x-3)(x+3)}{(x+5)^2} \]<br /><br />Next, cancel out the common factor \((x-3)\):<br /><br />\[ \frac{2}{\cancel{(x-3)}^2} \times \frac{\cancel{(x-3)}(x+3)}{(x+5)^2} \]<br /><br />This further simplifies to:<br /><br />\[ \frac{2(x+3)}{(x-3)(x+5)^2} \]<br /><br />Finally, distribute the terms in the denominator:<br /><br />\[ \frac{2(x+3)}{(x-3)(x+5)^2} = \frac{2(x+3)}{(x-3)(x+5)^2} \]<br /><br />Thus, the simplified form of the given expression is:<br /><br />\[ \boxed{\frac{2(x+3)}{(x-3)(x+5)^2}} \]